题意：首先给你个w，表示几组货物，然后给你n以及n个数表示价格，10-价格是利润

每次只能取前i个，求最多的利润对应几个货物，不超过10个解

思路：首先是简单的计算最大的利润，然后就是储存所有的解，然后深搜出所有的可能

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          using namespace std; const int MAXN = 100; int w, val[MAXN][MAXN]; set
         
           ans; vector
          
            p[MAXN]; void cal(int n, int sum) { if (n == w) { ans.insert(sum); return; } for (int i = 0; i < p[n].size(); i++) cal(n+1, sum+p[n][i]); } void solve() { ans.clear(); int valMax[MAXN], valSum, pro; valSum = 0; memset(valMax, 0, sizeof(valMax)); for (int i = 0; i < w; i++) { p[i].clear(); pro = 0; for (int j = 1; j <= val[i][0]; j++) { pro += 10 - val[i][j]; valMax[i] = max(valMax[i], pro); } valSum += valMax[i]; if (valMax[i] == 0) p[i].push_back(0); pro = 0; for (int j = 1; j <= val[i][0]; j++) { pro += 10 - val[i][j]; if (valMax[i] == pro) p[i].push_back(j); } } cal(0, 0); printf("Maximum profit is %d.\n", valSum); printf("Number of pruls to buy:"); int cnt = 0; for (set
           
            ::iterator i = ans.begin(); i != ans.end() && cnt != 10; i++, cnt++) printf(" %d", *i); printf("\n"); } int main() { int t = 0; while (scanf("%d", &w) != EOF && w) { int b; for (int i = 0; i < w; i++) { scanf("%d", &val[i][0]); for (int j = 1; j <= val[i][0]; j++) scanf("%d", &val[i][j]); } if (t) printf("\n"); printf("Workyards %d\n", ++t); solve(); } return 0; }